False
Steps of construction
Draw a line segment BC with suitable length
Taking B and C as centers draw two arcs of suitable radii intersecting each other at A
Join BA and CA. Δ ABC is the required triangle.
From B draw any ray BX downwards making an acute angle CBX.
Locate seven points B1,B2,B3....B7 on BX such that BB1=B1B2=B2B3=B3B4=B4B5=B5B6=B6B7.
Join B3C and form B7 draw a line B7C’ || B3C intersecting the extended line segment BC at C’.
From point C’ draw C’A || CA intersecting the extended line segment BA at A’. then Δ A’BC’ is the required triangle whose sides are of the corresponding sides of ΔABC. Given that, segment B6C’ is drawn parallel to B3C. But form our construction is never possible that segment B6C’ is parallel to B3C because the similar triangle A’BC’ has its sides of the corresponding sides of triangle ABC. So . B7C’ is parallel to B3C.